Note that if there is signal in any of these controls, it is likely not a problem unless there is a lot.

Note that if there is signal in any of these controls, it is likely not a problem unless there is a lot.

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For example, if an NAC probing for 18S showed a signal (Ct) of 31, while your samples showed 22, this represents a difference in 31-22=9 cycles of PCR. 9 cycles of PCR is equal to a 2^9=512 products. Said differently, during 9 cycles of PCR, the number of products generated increases 512-fold. So in the NAC sample, although it is contributing signal to the sample, it is only providing 1/512=0.002 or 0.2% of the signal, which is negligible.

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For example, if an NAC probing for 18S showed a signal (Ct) of 31, while your samples showed 22, this represents a difference in 31-22=9 cycles of PCR. 9 cycles of PCR is equal to a 2^9=512 products. So during 9 cycles of PCR, the number of products generated increases 512-fold. In the NAC sample, although it is contributing signal to the sample, it is only providing 1/512=0.002 or 0.2% of the signal, which seems negligible. If ANYONE CAN VERIFY THIS OH SCIENTIFIC WORLD...Please email me. I believe this is correct but I am not completely certain (I am biology, not math).

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Looked at another way, the amount of product (and therefore signal) after 22 cycles

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Revision as of 16:29, 5 November 2012

I typically use 10 ng cDNA/Taqman reaction and I deliver 5 μL, so you want a cDNA concentration of 2 ng/μL. For this example I am going to pretend I have 2 samples (X and Y as well as controls NTC and NAC) and I want to probe for 18S, IL-6, and OSM. The 96-well template will look something like this:

Row A will probe for 18S (orange), B will probe for IL-6 (green) and C will probe for OSM (purple). So these are the probes you will include in those wells.

Use this EXCEL TEMPLATE to analyze the data. The method I use involves relating IL-6/OSM expression to 18S. The values for this relative expression is then made to equal 1.00 for the control group and all other treatments are compared to this calibrated value. Essentially, the control group is considered the baseline, and treatment groups are expressed as increases/decreases in fold relative to this baseline, in terms of IL-6/OSM expression (in this example).

Please note that it is necessary to include PCR controls while doing this. Here, these controls are:

NAC: No Amplification Control (no reverse transcription - assesses for genomic DNA that can act as template)
This involves generating an RT reaction that contains everything from a representative sample from a single experiment
(experiments done at different times, in different cells, or if the RNA was isolated at a separate time - all need a single
representative NAC). So make a tube WITHOUT RTase, while including template and all other reagents. This verifies for the
presence of DNA that may be contributing signal.

NTCRT: No Template Control (Reverse Transcription) - assesses for contaminating DNA/RNA that can act as template within RT reagents
This involves putting together a tube during RT that contains everything but template. This only needs to be performed
once per Taqman run and must be done everytime Taqman is done. Evaluates whether or not there is contaminating DNA/RNA in
the RT reagents that is giving signal.

NTCPCR: No Template Control (PCR) - assesses for contaminating DNA/RNA that can act as template within PCR reagents
This involves preparing a tube for Taqman that contains everything but template. This needs to be performed once per
Taqman run and must be done everytime Taqman is done.Evaluates whether or not there is contaminating DNA in the Taqman
reagents that is giving signal.

Note that if there is signal in any of these controls, it is likely not a problem unless there is a lot.
For example, if an NAC probing for 18S showed a signal (Ct) of 31, while your samples showed 22, this represents a difference in 31-22=9 cycles of PCR. 9 cycles of PCR is equal to a 2^9=512 products. So during 9 cycles of PCR, the number of products generated increases 512-fold. In the NAC sample, although it is contributing signal to the sample, it is only providing 1/512=0.002 or 0.2% of the signal, which seems negligible. If ANYONE CAN VERIFY THIS OH SCIENTIFIC WORLD...Please email me. I believe this is correct but I am not completely certain (I am biology, not math).